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This is mainly a request for a straightforward reference (preferably at textbook level). The question comes up while responding to a question raised by non-specialists in finite group representations.

Let $G$ be a finite group and let $F \subset E$ be finite fields, say with $E$ a splitting field for $G$ but perhaps $F$ not. Starting with a simple module $M$ of dimension $d$ for the group algebra $E[G]$, consider the $F[G]$-module $N$ obtained by "restriction of scalars" from $M$: here $M$ as a vector space over $F$ typically has larger dimension related to $d$ while the representing matrices for $G$ over $F$ involve a rewriting of the matrices over $E$. While this is elementary in principle, it's a bit complicated to write down in practice. In the process, one could chart a precise relationship between simple modules over the field $F$ (maybe not a splitting field) and those over $E$.

Is this written down clearly for non-specialists?

Going the other way, from a module over an arbitrary field $F$ to one over a splitting field $E$, is a more standard topic in textbooks about which much can be said in prime characteristic as well as in characteristic 0 (where the Schur index becomes a leading idea). See for example the extensive discussion in Chapter 9 "Changing the field" in Character Theory of Finite Groups by I.M. Isaacs along with the following chapter. But I don't recall seeing comparable detail on restriction of scalars. The question I've been discussing with colleagues isn't especially deep or advanced, but a clear reference would help.

ADDED: To make the situation more precise, it may be enough (following Geoff) to assume that $F$ is not a splitting field for the given simple $E[G]$-module $M$ while $E$ is a minimal splitting field for it. (Or you might just take $E$ to be a minimal splitting field for $G$.) Then if $[E:F] = n$, the restricted $F[G]$-module $N$ has dimension $nd$ over $F$. In this situation one expects $N$ to be simple for the smaller group algebra, which in turn yields an $nd$-dimensional module over $E$ after tensoring, etc. The fields involved being finite, there is no Schur index complication and the Galois group of $E/F$ is easy to work with. So the matrix version over $F$ of the given representation over $E$ should be readily described in block form. With luck, this is all I've been asked to explain in special cases, but it would be reassuring to see a concise published version which I couldn't readily extract from books I've seen.

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Just to make sure I understand what you mean: if $M$ is a simple module for $E[G]$, and $N$ is the $F[G]$-module obtained by "restriction scalars", then $\dim_F N = [E:F] \dim_E M$, right? I didn't quite know what you meant by "...typically has larger dimension..." -- it seems that it has larger dimension whenever $E \ne F$. –  George McNinch Mar 14 '12 at 0:41
    
Maybe you mean: the composition factors $S$ of $N$ as $F[G]$ module can have $\dim_F S > \dim_E M$. –  George McNinch Mar 14 '12 at 0:44
    
@George: My initial formulation was a little loose, in keeping with the somewhat convoluted quewtions I was being asked. So I've tried to tighten the assumptiuons. (My language "restriction of scalars" isn't precise, but imitates terminology used by Weil and others for algebraic group structure relative to field extensions. It's probably not helpful here.) –  Jim Humphreys Mar 14 '12 at 22:45

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I'm sure you have worked all this out, but the representation over $E$ can be realised over the extension of the prime subfield generated by the traces of the group elements (boiling down to the absence of Schur indices over finite fields, and ultimately to the fact that finite divisions rings are fields). If you assume that $E$ is that minimal field then the unique simple $FG$-module $M$ which has the given one as simple summand (after extension of scalars to $E$) has dimension $[E:F]$ times the original dimension, as George suggests, and is the sum of $[E:F]$ Galois conjugates of the original one. Furthermore, $E$ is the ring of $FG$-endomorphisms of $M$. If the field $E$ is not this minimal field, it seems less obvious to me how to realise the representation over the subfield generated by the traces, and I suppose this may be the real issue you need to resolve. Note added later: To be even more explicit in the case where $E$ is minimal, we may write $E = F[\alpha]$ where $\alpha$ is a generator for the multiplicative group of $E.$ Then $\alpha$ has minimum polynomial $m(x)$ of degree $n = [E:F]$ over $F.$ Let $C$ be the companion matrix for $m(x)$ which is an $n \times n$ matrix with entries in $F.$ Then whenever $\alpha^{j}$ appears as an entry of a matrix in the given representation over $E,$ replace it by the $n \times n$ block $C^{j}$ (and put an $n \times n$ block of zeros whnever $0$ occurs in the original matrix representation). This gives the representation over $F$ of dimension $[E:F]$ times the dimension of the original representation, which is unique up to equivalence.

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Thanks for the clarifications, some of which I've understood but not in an organized way; at least your first sentence had been clear to me, but then my sense of the possibilities got fuzzy. I'll add a little more to my question, but meanwhile I'm still hoping to find a source in the literature which will be readable by other non-specialists. (Even more non-specialized in these matters than I am, but raising reasonable questions in small cases.) –  Jim Humphreys Mar 14 '12 at 20:08
    
I'm afraid I can't think of any obvious reference where these things are treated explicitly, though such references must exist. –  Geoff Robinson Mar 14 '12 at 20:52
    
Thanks also for the added details. (I trust all of this will be presented with optimal efficiency in your forthcoming textbook ...?) In spite of your optimistic "such references must exist", I'm fairly convinced that none of the standard books tells quite the full story here; but I think I can piece it together better now. –  Jim Humphreys Mar 16 '12 at 16:20
    
Thanks Jim. I'll try and remember to include it when I write my book :) –  Geoff Robinson Mar 16 '12 at 18:04

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